The sum to infinity of the following series 2 | vedclass

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(C) The given series is $S = 2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots \infty$. We can rea

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$7/2$

Vedclass · Answer

(C) The given series is $S = 2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots \infty$. We can rearrange the terms as: $S = (1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \infty) + (1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots \infty)$. Both parts are infinite geometric series with the sum formula $S_{\infty} = \frac{a}{1-r}$. For the first series,$a = 1$ and $r = 1/2$,so $S_1 = \frac{1}{1 - 1/2} = 2$. For the second series,$a = 1$ and $r = 1/3$,so $S_2 = \frac{1}{1 - 1/3} = \frac{1}{2/3} = 3/2$. Thus,the total sum is $S = S_1 + S_2 = 2 + 3/2 = 7/2$.

The sum to infinity of the following series $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots$ is:

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